Calculate the pH when 64.0 mL of 0.150 M KOH is mixed with 20.0 mL of 0.300 M HBrO (Ka = 2.5 × 10⁻⁹)

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Chemistry Ngọc Hoa 5 months 2021-09-05T06:18:47+00:00 2021-09-05T06:18:47+00:00 1 Answers 8 views 0

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Captcha* Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )

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thanhha · Answer 1 · 2021-09-05T06:19:56+00:00

thanhha

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2021-09-05T06:19:56+00:00 September 5, 2021 at 6:19 am

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Answer:

The answer is “12.06“

Explanation:

Given:

$M(HBrO) = 0.3\ M\\\\V(HBrO) = 20 \ mL\\\\M(KOH) = 0.15 \ M\\\\V(KOH) = 64 \ mL$

$\to mol(HBrO) = M(HBrO) \times V(HBrO) = 0.3 M \times 20 mL = 6 \ mmol\\\\\to mol(KOH) = M(KOH) \times V(KOH)= 0.15 M \times 64 mL = 9.6 mmol$

6 mmol of both will react

excess KOH remaining $= 3.15 \ mmol$

Volume $= 20 + 64 = 84 \ mL$

$[OH^{-}] = \frac{ 9.6 \ mmol}{84\ mL} = 0.01142\ M$

use:

$pOH = -\log [OH^-]$

$= -\log (1.142\times 10^{-2})\\\\= 1.94$

use:

$PH = 14 - pOH$

$= 14 - 1.94\\\\= 12.06$